A body is projected up with a velocity equal to `3//4th` of the escape velocity from the surface of the earth. The height it reaches is (Radius of the earth is `R`)

Options

A. `(10 R)/(9)`
B. `(9R)/(7)`
C. `(9R)/(8)`
D. `(10R)/(3)`

Answer

B

Explanation

Escape velocity, `v_e = sqrt((2GM )/(R))`
Velocity of projection, `v = 3/4 v_e = 3/4 sqrt ((2GM)/( R))`
The total energy of the body when it is projected is
`E_i = KE + PE`
` = 1/2mv^2 - (GmM)/(R ) = 1/2 m xx 9/16 xx (2GM)/( R) - (GmM)/( R)`
`= 9/16 (GmM)/( R) - (GmM)/( R) = - 7/16 (GmM)/( R)`
Let h be the maximum height attained by the body. The distance of the body from the centre of the earth is r = R + h. At this height, the total energy of the body is
`E_f =0 - (GmM)/( r) = (GmM)/( r) = - (GmM)/((R+ h))`
From the law of conservation of energy, `E_i = E_f`
` therefore - 7/16 (GmM)/( R) = - (GmM)/((R+ h))`
`7(R + h) = 16R " or " 7h = 9R " or " h = (9R)/(7)`